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Home/ Questions/Q 7669951
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T15:42:42+00:00 2026-05-31T15:42:42+00:00

I have three difference examples mention below. I don’t understand why ex1 has same

  • 0

I have three difference examples mention below. I don’t understand why ex1 has same output for ex2 and differ output for ex3, also why ex2 is not the same as ex3 where I just make a creation in another line!!

ex1

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int x=2;
    int *y;
    y = &x;
    printf("value: %d\n", *y);
    printf("address: %d\n", y);
    return EXIT_SUCCESS;
}

output

value: 2
address: 2686744

ex2

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int x=2;
    int *y = &x;
    printf("value: %d\n", *y);
    printf("address: %d\n", y);
    return EXIT_SUCCESS;
}

output

value: 2
address: 2686744

ex3

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int x=2;
    int *y;
    *y = &x;
    printf("value: %d\n", *y);
    printf("address: %d\n", y);
    return EXIT_SUCCESS;
}

output

value: 2686744
address: 2130567168

I HAVE BIG MISUNDERSTANDING OF POINTERS WHEN I THINK STAR MUST BECOME WITH (y) NOT (int)
AND I FIGURE OUT THAT STAR WITH (int) NOT (y) (^_^)
NOW EVERYTHING IS CLEAR FOR ME… THANKS FOR ALL YOUR ANSWERS

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T15:42:43+00:00Added an answer on May 31, 2026 at 3:42 pm

    In example 3, you first declare a pointer:

    int *y;
    

    and then you say that the int value of *y is the address of x.

    That’s because with the declaration int *y you have:

    • y is of type int *
    • *y is of type int.

    So, the right lines of code in example 3 should be:

    int *y;
    y = &x;
    
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